Let N be the smallest positive integer that leaves a remainder of 3 when divided by 11, a … — State PSC Exam

Let $N$ be the smallest positive integer that leaves a remainder of $3$ when divided by $11$ , a remainder of $5$ when divided by $13$ , and is exactly divisible by $7$ . What is the remainder when $N^4 + N^2 + 1$ is divided by $17$ ?

Answer: 3

Answer

The remainder when

N^4 + N^2 + 1

is divided by

17

is 3.

Using the Chinese Remainder Theorem, the smallest integer meeting all divisibility constraints is determined to be

N = 707

. Reducing

707

modulo

17

gives

10

. Substituting

10

into the expression

N^4 + N^2 + 1

modulo

17

yields

(10^4 + 10^2 + 1) \pmod{17} = (4 + 15 + 1) \pmod{17} = 20 \pmod{17} = 3

Step-by-Step Solution

Express the given conditions as a system of linear congruences

N \equiv 3 \pmod{11}

N \equiv 5 \pmod{13}

, and

N \equiv 0 \pmod{7}

Translates the problem statement into standard modular arithmetic notation.

Solve the simultaneous congruences for modulo 11 and modulo 13

Let

N = 11k + 3

. Substituting into

11k + 3 \equiv 5 \pmod{13}

gives

11k \equiv 2 \pmod{13} \implies -2k \equiv 2 \pmod{13} \implies k \equiv 12 \pmod{13}

. Thus,

N = 11(13m + 12) + 3 = 143m + 135

Combines two modular constraints into a single general expression using the Chinese Remainder Theorem principle.

Apply the divisibility condition for 7 to determine the smallest value of

N

143m + 135 \equiv 0 \pmod{7} \implies 3m + 2 \equiv 0 \pmod{7} \implies 3m \equiv 5 \pmod{7} \implies m \equiv 4 \pmod{7}

. For the smallest positive integer, set

m = 4

, so

N = 143(4) + 135 = 707

Finds the specific smallest integer

N

that satisfies all three divisor requirements.

Reduce

N

modulo 17

707 = 17 \times 41 + 10 \implies N \equiv 10 \pmod{17}

Simplifies the base before evaluating higher powers.

Calculate the polynomial expression

(N^4 + N^2 + 1) \pmod{17}

N^2 \equiv 10^2 = 100 \equiv 15 \equiv -2 \pmod{17}

, and

N^4 \equiv (-2)^2 = 4 \pmod{17}

. Therefore,

N^4 + N^2 + 1 \equiv 4 + 15 + 1 = 20 \equiv 3 \pmod{17}

Uses modular exponentiation properties to efficiently evaluate the final remainder.

Key Concept

Chinese Remainder Theorem and Modular Exponentiation

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